а) Для уравнения \(5x^2 - 11x + 2 = 0\), \(a = 5\), \(b = -11\), \(c = 2\).
\(D = (-11)^2 - 4 cdot 5 cdot 2 = 121 - 40 = 81\).
\(x_1 = \frac{-(-11) + \sqrt{81}}{2 cdot 5} = \frac{11 + 9}{10} = \frac{20}{10} = 2\)
\(x_2 = \frac{-(-11) - \sqrt{81}}{2 cdot 5} = \frac{11 - 9}{10} = \frac{2}{10} = \frac{1}{5}\)
б) Для уравнения \(2p^2 + 7p - 30 = 0\), \(a = 2\), \(b = 7\), \(c = -30\).
\(D = 7^2 - 4 cdot 2 cdot (-30) = 49 + 240 = 289\).
\(p_1 = \frac{-7 + \sqrt{289}}{2 cdot 2} = \frac{-7 + 17}{4} = \frac{10}{4} = \frac{5}{2}\)
\(p_2 = \frac{-7 - \sqrt{289}}{2 cdot 2} = \frac{-7 - 17}{4} = \frac{-24}{4} = -6\)
в) Для уравнения \(9y^2 - 30y + 25 = 0\), \(a = 9\), \(b = -30\), \(c = 25\).
\(D = (-30)^2 - 4 cdot 9 cdot 25 = 900 - 900 = 0\). Так как \(D = 0\), уравнение имеет 1 корень.
\(y = \frac{-(-30) + \sqrt{0}}{2 cdot 9} = \frac{30}{18} = \frac{5}{3}\)
**Ответ:**
а) x₁ = 2, x₂ = 1/5
б) p₁ = 5/2, p₂ = -6
в) y = 5/3